Chapter 3, Problem 5P

A fish swimming in a horizontal plane has velocity ${\stackrel{\to }{v}}_i=(4.00\hat{i}+1.00\hat{j})\text{m/s}$ at a point in the ocean where the position relative to a certain rock is ${\stackrel{\to }{r}}_i=(10.0\hat{i}-4.00\hat{j})\text{m}$. After the fish swims with constant acceleration for 20.0 s, its velocity is $\stackrel{\to }{v}=(20.0\hat{i}-5.00\hat{j})\text{m/s}$. (a) What are the components of the acceleration of the fish? (b) What is the direction of its acceleration with respect to unit vector $\hat{i}$? (c) If the fish maintains constant acceleration, where is it at t = 25.0 s and in what direction is it moving?

To determine

The components of the acceleration of the fish .

Answer to Problem 5P

The components of the acceleration of the fish is $\underset{\_}{0.800{\text{m/s}}^{\text{2}}}$ and $\underset{\_}{-0.300{\text{m/s}}^{\text{2}}}$.

Explanation of Solution

Write the expression for the $x$ component of the acceleration of the fish.

  $a_x=\frac{\Delta v_x}{\Delta t}$        (I)

Here, $a_x$ is the $x$ component of the acceleration of the fish, $\Delta v_x$ is the change in the $x$ component of the velocity of the fish and $\Delta t$ is the time taken.

Write the expression for the $y$ component of the acceleration of the fish.

  $a_y=\frac{\Delta v_y}{\Delta t}$        (II)

Here, $a_y$ is the $y$ component of the acceleration of the fish, $\Delta v_y$ is the change in the $y$ component of the velocity of the fish and $\Delta t$ is the time taken.

Conclusion:

Substitute $20.0\text{m/s}-\text{4}\text{.00}\text{m/s}$ for $\Delta v_x$ and $20.0\text{s}$ for $\Delta t$ in (I) to find $a_x$.

    $\begin{array}{c}{a}_x=\frac{20.0\text{m/s}-\text{4}\text{.00}\text{m/s}}{20.0\text{s}}\\ =0.800{\text{m/s}}^{\text{2}}\end{array}$

Substitute $-5.0\text{m/s}-1.\text{00}\text{m/s}$ for $\Delta v_x$ and $20.0\text{s}$ for $\Delta t$ in (II) to find $a_y$.

    $\begin{array}{c}{a}_y=\frac{-5.0\text{m/s}-1.\text{00}\text{m/s}}{20.0\text{s}}\\ =-0.300{\text{m/s}}^{\text{2}}\end{array}$

Therefore, the components of the acceleration of the fish is $\underset{\_}{0.800{\text{m/s}}^{\text{2}}}$ and $\underset{\_}{-0.300{\text{m/s}}^{\text{2}}}$.

To determine

The angle of its acceleration with respect to unitvector $\widehat{\text{i}}$.

Answer to Problem 5P

The angle of its acceleration with respect to unit vector $\widehat{\text{i}}$ is ${339}^{\circ }\text{from}+x\text{axis}$.

Explanation of Solution

Write the expression for the angle of its acceleration with respect to unit vector $\widehat{\text{i}}$.

  $\theta ={\tan }^{-1}\left(\frac{a_y}{a_x}\right)$        (III)

Here, $\theta$ is the angle of its acceleration with respect to unit vector $\widehat{\text{i}}$.

Conclusion:

Substitute $0.800{\text{m/s}}^{\text{2}}$ for $a_x$ and $-0.300{\text{m/s}}^{\text{2}}$ for $a_y$ in (III) to find $\theta$,

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-0.300{\text{m/s}}^{\text{2}}}{0.800{\text{m/s}}^{\text{2}}}\right)\\ =-{20.6}^{\circ }\\ ={339}^{\circ }\end{array}$

Therefore, the angle of its acceleration with respect to unit vector $\widehat{\text{i}}$ is ${339}^{\circ }\text{from}+x\text{axis}$.

To determine

The angle at time $t=25.0\text{s}$ when the fish moving in constant acceleration.

Answer to Problem 5P

The angle at time $t=25.0\text{s}$ when the fish moving in constant acceleration is $-{15.2}^{\circ }$.

Explanation of Solution

Write the expression for the $x$ component of the position at time $t$ .

    $x_f=x_i+v_{xi}t+\frac{1}{2}{a}_x{t}^2$        (IV)

Here, $x_f$ is the final position in $x$ component, $x_i$ is initial position in $x$ component and $v_{xi}$ is the initial speed in $x$ component.

Write the expression for the $y$ component of the position at time $t$.

    $y_f=y_i+v_{yi}t+\frac{1}{2}{a}_y{t}^2$        (V)

Here, $y_f$ is the final position in $y$ component, $y_i$ is initial position in $y$ component and $v_{yi}$ is the initial speed in $y$ component.

Write the expression for the $x$ component of the final velocity of the fish.

    $v_{xf}=v_{xi}+a_{x}t$        (VI)

Here, $v_{xf}$ is the $x$ component of the final velocity of the fish, $v_{xi}$ is the $x$ component of the initial velocity of the fish and $a_x$ is the $x$ component of the acceleration of the fish.

Write the expression for the $y$ component of the final velocity of the fish.

    $v_{yf}=v_{yi}+a_{y}t$        (VII)

Here, $v_{yf}$ is the $x$ component of the final velocity of the fish, $v_{yi}$ is the $y$ component of the initial velocity of the fish and $a_y$ is the $y$ component of the acceleration of the fish.

Write the angle at time $t$ when the fish moving in constant acceleration.

    $\theta ={\tan }^{-1}\left(\frac{v_{yf}}{v_{xf}}\right)$        (VIII)

Here, $v_{xf}$ is the velocity of the fish in the $x$ component and $v_{yf}$ is the velocity of the fishin the $y$ component.

Conclusion:

Substitute $10.0\text{m}$ for $x_i$ , $4.00\text{m/s}$ for $v_{xi}$ , $0.800{\text{m/s}}^{\text{2}}$ for $a_x$ and $25.0\text{s}$ for $t$ in (IV) to find $x_f$,

    $\begin{array}{c}{x}_f=10.0\text{m}+\left(4.00\text{m/s}\right)\left(25.0\text{s}\right)+\frac{1}{2}\left(0.800{\text{m/s}}^{\text{2}}\right){\left(25.0\text{s}\right)}^2\\ =360\text{m}\end{array}$

Substitute $-4.0\text{m}$ for $y_i$ , $1.00\text{m/s}$ for $v_{yi}$ , $-0.300{\text{m/s}}^{\text{2}}$ for $a_y$ and $25.0\text{s}$ for $t$ in (V) to find $y_f$,

    $\begin{array}{c}{y}_f=-4.0\text{m}+\left(1.00\text{m/s}\right)\left(25.0\text{s}\right)+\frac{1}{2}\left(-0.300{\text{m/s}}^{\text{2}}\right){\left(25.0\text{s}\right)}^2\\ =-72.7\text{m}\end{array}$

Substitute $4.00\text{m/s}$ for $v_{xi}$ , $0.800{\text{m/s}}^{\text{2}}$ for $a_x$ and $25.0\text{s}$ for $t$ in (VI) to find $v_{xf}$,

    $\begin{array}{c}{v}_{xf}=4.00\text{m/s}+\left(0.800{\text{m/s}}^{\text{2}}\right)\left(25.0\text{s}\right)\\ =24\text{m/s}\end{array}$

Substitute $1.00\text{m/s}$ for $v_{yi}$ , $-0.300{\text{m/s}}^{\text{2}}$ for $a_y$ and $25.0\text{s}$ for $t$ in (VII) to find $v_{yf}$,

    $\begin{array}{c}{v}_{yf}=1.00\text{m/s}+\left(-0.300{\text{m/s}}^{\text{2}}\right)\left(25.0\text{s}\right)\\ =-6.50\text{m/s}\end{array}$

Substitute $-6.50\text{m/s}$ for $v_{yf}$ and $24\text{m/s}$ for $v_{xf}$ in (VIII) to find $\theta$ ,

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-6.50\text{m/s}}{24\text{m/s}}\right)\\ =-{15.2}^{\circ }\end{array}$

Therefore, the angle at time $t=25.0\text{s}$ when the fish moving in constant acceleration is $-{15.2}^{\circ }$.